The line y = mx bisects the area enclosed by the curve y =1+4 x - x 2 the lines x =0, x =(3/2) y=0. Then the value of m is :

Question

The line y = mx bisects the area enclosed by the curve y =1+4 x - x 2 the lines x =0, x =(3/2) y=0. Then the value of m is : (A) (13/6) (B) (6/13) (C) (3/2) (D) 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/653a0984ff5b19b78962ec7f0ae32776-.png" /> 2 ⋅ (1/2)((3/2) ⋅ (3 m /2))=∫ limits03 / 2(1+4 x - x 2) dx .(9 m /4)= x +2 x 2-( x 3/3)]03 / 2=(3/2)+(9/2)-(27/8) ⋅ (1/3)=6-(9/8)=(39/8) m =(13/6)

Q. The line $y = m x$ bisects the area enclosed by the curve $y = 1 + 4 x - x^{2} &$ the lines $x = 0, x = \frac{3}{2}$ $& y = 0$ . Then the value of $m$ is :

$\frac{13}{6}$

$\frac{6}{13}$

$\frac{3}{2}$

4

$2 \cdot \frac{1}{2} (\frac{3}{2} \cdot \frac{3 m}{2}) = 0 \int 3/2 (1 + 4 x - x^{2}) d x$
$\frac{9 m}{4} = x + 2 x^{2} - \frac{x ^{3}}{3}]_{0}^{3/2} = \frac{3}{2} + \frac{9}{2} - \frac{27}{8} \cdot \frac{1}{3} = 6 - \frac{9}{8} = \frac{39}{8}$
$m = \frac{13}{6}$

Q. The line y=mx bisects the area enclosed by the curve y=1+4x−x2& the lines x=0,x=23​ &y=0. Then the value of m is :

613​

136​

23​

4

2⋅21​(23​⋅23m​)=0∫3/2​(1+4x−x2)dx 49m​=x+2x2−3x3​]03/2​=23​+29​−827​⋅31​=6−89​=839​ m=613​

Q. The line $y = m x$ bisects the area enclosed by the curve $y = 1 + 4 x - x^{2} &$ the lines $x = 0, x = \frac{3}{2}$ $& y = 0$ . Then the value of $m$ is :

$\frac{13}{6}$

$\frac{6}{13}$

$\frac{3}{2}$

$2 \cdot \frac{1}{2} (\frac{3}{2} \cdot \frac{3 m}{2}) = 0 \int 3/2 (1 + 4 x - x^{2}) d x$
$\frac{9 m}{4} = x + 2 x^{2} - \frac{x ^{3}}{3}]_{0}^{3/2} = \frac{3}{2} + \frac{9}{2} - \frac{27}{8} \cdot \frac{1}{3} = 6 - \frac{9}{8} = \frac{39}{8}$
$m = \frac{13}{6}$